Suppose `0lt thetalt(pi)/(2)` if the eccentricity of the hyperbola `x^(2)-y^(2) cosec^(2) theta=5` is `sqrt(7)` times the eccentricity of the ellipse `x^(2) cosec^(2) theta+y^(2)=5` then `theta` is equal to ________

Answer

To solve the problem, we need to find the value of \(\theta\) given the relationship between the eccentricities of a hyperbola and an ellipse. Let's break it down step by step.

### Step 1: Identify the hyperbola and ellipse equations
The hyperbola is given by:
\[ x^2 - y^2 \csc^2 \theta = 5 \]
We can rewrite this in standard form:
\[ \frac{x^2}{5} - \frac{y^2}{5 \sin^2 \theta} = 1 \]
From this, we can identify:
- \(a^2 = 5\)
- \(b^2 = 5 \sin^2 \theta\)

### Step 2: Calculate the eccentricity of the hyperbola
The eccentricity \(E_1\) of a hyperbola is given by:
\[ E_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5 \sin^2 \theta}{5}} = \sqrt{1 + \sin^2 \theta} \]

### Step 3: Identify the ellipse equation
The ellipse is given by:
\[ x^2 \csc^2 \theta + y^2 = 5 \]
We can rewrite this in standard form:
\[ \frac{x^2}{5 \sin^2 \theta} + \frac{y^2}{5} = 1 \]
From this, we can identify:
- \(a^2 = 5\)
- \(b^2 = 5 \sin^2 \theta\)

### Step 4: Calculate the eccentricity of the ellipse
The eccentricity \(E_2\) of an ellipse is given by:
\[ E_2 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5 \sin^2 \theta}{5}} = \sqrt{1 - \sin^2 \theta} = \cos \theta \]

### Step 5: Set up the relationship between the eccentricities
According to the problem, the eccentricity of the hyperbola is \(\sqrt{7}\) times the eccentricity of the ellipse:
\[ E_1 = \sqrt{7} E_2 \]
Substituting the expressions for \(E_1\) and \(E_2\):
\[ \sqrt{1 + \sin^2 \theta} = \sqrt{7} \cos \theta \]

### Step 6: Square both sides
Squaring both sides gives:
\[ 1 + \sin^2 \theta = 7 \cos^2 \theta \]

### Step 7: Use the identity \(\sin^2 \theta + \cos^2 \theta = 1\)
We can replace \(\sin^2 \theta\) with \(1 - \cos^2 \theta\):
\[ 1 + (1 - \cos^2 \theta) = 7 \cos^2 \theta \]
This simplifies to:
\[ 2 = 8 \cos^2 \theta \]

### Step 8: Solve for \(\cos^2 \theta\)
Dividing both sides by 8:
\[ \cos^2 \theta = \frac{1}{4} \]

### Step 9: Find \(\cos \theta\)
Taking the square root:
\[ \cos \theta = \frac{1}{2} \quad \text{(since \(0 < \theta < \frac{\pi}{2}\))} \]

### Step 10: Determine \(\theta\)
The angle \(\theta\) corresponding to \(\cos \theta = \frac{1}{2}\) is:
\[ \theta = \frac{\pi}{3} \]

Thus, the value of \(\theta\) is:
\[ \theta = \frac{\pi}{3} \]